BMAD 62 Report post Posted May 9, 2013 The sum of three numbers is 6, the sum of their squares is 8, and the sum of their cubes is 5. What is the sum of their fourth powers? Share this post Link to post Share on other sites

0 bonanova 76 Report post Posted May 11, 2013 Interesting result! I don't solve cubics. I am retired, and I don't have to do things like that. Or Bayes formulas, at least until I learn to understand them. Newton-Raphson, on the other hand is a convenient and generalsledge-hammer tool, and it's fairly simple to program. I did a manual search that convinced me the numbers are not integers, and they are not (all) real. NR tells me that are of the forma = realb = x + iyc = x - iy With values ofa = 2.67770427706x = 1.66114786147y = 1.53116371279These values gave 6 8 and 5 for sums squares and cubes, within 10^{-8}.The sum of their 4th powers? Zero!I found an alternative and interesting triplet that someone might want to solve. It may be quite easy because [white spoiler] a, x and y are all integers [/white spoiler]. Hint: to see the spoiler, just select it.a + b + c = 4a^{2}+ b^{2}+ c^{2} = 4a^{3}+ b^{3}+ c^{3} = 4a^{4}+ b^{4}+ c^{4} = ___? Share this post Link to post Share on other sites

0 dark_magician_92 4 Report post Posted May 10, 2013 Any Hint for me? Share this post Link to post Share on other sites

0 BMAD 62 Report post Posted May 10, 2013 Let the numbers be a, b, and c. Then we have a + b + c = 6 a^{2} + b^{2} + c^{2} = 8 a^{3} + b^{3} + c^{3} = 5 find the monic cubic equation Share this post Link to post Share on other sites

0 TimeSpaceLightForce 11 Report post Posted May 11, 2013 Let the numbers be a, b, and c. Then we have a + b + c = 6 a^{2} + b^{2} + c^{2} = 8 a^{3} + b^{3} + c^{3} = 5 find the monic cubic equation a^{3} - a^{2} - a + 1 + b^{3} - b^{2} - b + 1 + c^{3 }- c^{2} - c + 1 = 5 - 8 - 6 + 3 (a+1)(a-1)(a-1) + (b+1)(b-1)(b-1) + (c+1)(c-1)(c-1) = - 6 using excel: try for a & b [a] [b] [c=d^(1/3)] :sum 6 ? [=a^2] [=b^2] [=c^2] :sum 8 ? [=a^3] [=b^3] [d=5-a^3-b^3] :sum 5 ! Share this post Link to post Share on other sites

0 dark_magician_92 4 Report post Posted May 11, 2013 Interesting result! I don't solve cubics. I am retired, and I don't have to do things like that. Or Bayes formulas, at least until I learn to understand them. Newton-Raphson, on the other hand is a convenient and generalsledge-hammer tool, and it's fairly simple to program. I did a manual search that convinced me the numbers are not integers, and they are not (all) real. NR tells me that are of the forma = realb = x + iyc = x - iy With values ofa = 2.67770427706x = 1.66114786147y = 1.53116371279 These values gave 6 8 and 5 for sums squares and cubes, within 10^{-8}. The sum of their 4th powers? Zero! I found an alternative and interesting triplet that someone might want to solve. It may be quite easy because [white spoiler] a, x and y are all integers [/white spoiler]. Hint: to see the spoiler, just select it. a + b + c = 4a^{2}+ b^{2}+ c^{2} = 4a^{3}+ b^{3}+ c^{3} = 4a^{4}+ b^{4}+ c^{4} = ___? It appeared to me too that a,b,c all cant be real. If we could find the value of a*b*c, then the problem could be done. Can you please provide a link for newton raphson method (some easy explanation ) Share this post Link to post Share on other sites

0 BMAD 62 Report post Posted May 11, 2013 If cubic equation x^{3} − Ax^{2} + Bx − C = 0 has roots a, b, c, then, expanding (x − a)(x − b)(x − c), we find A = a + b + c B = ab + bc + ca C = abc Then B = ab + bc + ca = ½ [(a + b + c)^{2} − (a^{2} + b^{2} + c^{2})] = 14. Hence a, b, c are roots of x^{3} − 6x^{2} + 14x − C = 0, and we have a^{3} − 6a^{2} + 14a − C = 0 b^{3} − 6b^{2} + 14b − C = 0 c^{3} − 6c^{2} + 14c − C = 0 Adding, we have (a^{3} + b^{3} + c^{3}) − 6(a^{2} + b^{2} + c^{2}) + 14(a + b + c) − 3C = 5 − 6×8 + 14×6 − 3C = 0. Hence C = 41/3, and x^{3} − 6x^{2} + 14x − 41/3 = 0. Multiplying the polynomial by x, we have x^{4} − 6x^{3} + 14x^{2} − 41x/3 = 0. Then a^{4} − 6a^{3} + 14a^{2} − 41a/3 = 0 b^{4} − 6b^{3} + 14b^{2} − 41b/3 = 0 c^{4} − 6c^{3} + 14c^{2} − 41c/3 = 0 Adding, we have (a^{4} + b^{4} + c^{4}) − 6(a^{3} + b^{3} + c^{3}) + 14(a^{2} + b^{2} + c^{2}) − 41(a + b + c)/3 = 0. Hence a^{4} + b^{4} + c^{4} = 6×5 − 14×8 + (41/3)×6 = 0. That is, the sum of the fourth powers of the numbers is 0. Share this post Link to post Share on other sites

0 dark_magician_92 4 Report post Posted May 11, 2013 If cubic equation x^{3} − Ax^{2} + Bx − C = 0 has roots a, b, c, then, expanding (x − a)(x − b)(x − c), we find A = a + b + c B = ab + bc + ca C = abc Then B = ab + bc + ca = ½ [(a + b + c)^{2} − (a^{2} + b^{2} + c^{2})] = 14. Hence a, b, c are roots of x^{3} − 6x^{2} + 14x − C = 0, and we have a^{3} − 6a^{2} + 14a − C = 0 b^{3} − 6b^{2} + 14b − C = 0 c^{3} − 6c^{2} + 14c − C = 0 Adding, we have (a^{3} + b^{3} + c^{3}) − 6(a^{2} + b^{2} + c^{2}) + 14(a + b + c) − 3C = 5 − 6×8 + 14×6 − 3C = 0. Hence C = 41/3, and x^{3} − 6x^{2} + 14x − 41/3 = 0. Multiplying the polynomial by x, we have x^{4} − 6x^{3} + 14x^{2} − 41x/3 = 0. Then a^{4} − 6a^{3} + 14a^{2} − 41a/3 = 0 b^{4} − 6b^{3} + 14b^{2} − 41b/3 = 0 c^{4} − 6c^{3} + 14c^{2} − 41c/3 = 0 Adding, we have (a^{4} + b^{4} + c^{4}) − 6(a^{3} + b^{3} + c^{3}) + 14(a^{2} + b^{2} + c^{2}) − 41(a + b + c)/3 = 0. Hence a^{4} + b^{4} + c^{4} = 6×5 − 14×8 + (41/3)×6 = 0. That is, the sum of the fourth powers of the numbers is 0. With this method we find abc=41/3. Now if we expand (a+b+c)^4 as (a+b+c)^2*(a+b+c)^2, then we get a^4+b^4+c^4=24*abc-328, which on substituting abc becomes = 0. Share this post Link to post Share on other sites

0 bonanova 76 Report post Posted May 11, 2013 dm92 asks for info on Newton method. Several links out there, this one is OK.http://m.youtube.com/watch?v=ER5B_YBFMJo&feature=related Successive approximations. You solve f(x)=0. You give an initial guess x0. Then x1 = x0 - f(x0)/f'(x0). Repeat. In several dimensions, like this problem with real numbers a x and y, the functions are f1 = 6-sum abc f2 = 8-sum2 abc f3 = 5-sum3 abc Calculate a 3x3 matrix M of partial derivatives of f123 wrt axy Guess axy, and repeat axy = axy - M^{-1} x f123 You can use delta's to approximate the derivatives if f is messy. Share this post Link to post Share on other sites

0 BMAD 62 Report post Posted May 11, 2013 Note that we did not need to actually calculate a, b, and c in order to determine the sum of their fourth powers. In fact, one of the numbers is real; the other two are complex conjugates as Bonanova points out; see below. The approximate values of the numbers are 2.67770, and 1.66115 ± 1.53116i. now in response to Bonanova: Generalization Using the above approach, we can show that if a + b + c = r a^{2} + b^{2} + c^{2} = s a^{3} + b^{3} + c^{3} = t then a, b, c are roots It then follows that of x^{3} − rx^{2} + ½(r^{2} − s)x + (½r(3s − r^{2}) − t)/3 = 0 a^{4} + b^{4} + c^{4}= 4rt/3 − ½s(r^{2} − s) + r^{2}(r^{2} − 3s)/6= (r^{4} − 6r^{2}s + 3s^{2} + 8rt)/6. Recurrence Relation Let f(n) = an + bn + cn, where n is a positive integer. Then, given the equation x^{3} − 6x^{2} + 14x − 41/3 = 0, we can multiply by x^{n} and sum over the three roots to yield the following recurrence relation: f(n+3) = 6f(n+2) − 14f(n+1) + (41/3)f(n). Successive applications of this formula allow us to calculate a^{5} + b^{5} + c^{5}, a^{6} + b^{6} + c^{6}, and so on. Share this post Link to post Share on other sites

The sum of three numbers is 6, the sum of their squares is 8, and the sum of their cubes is 5. What is the sum of their fourth powers?

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